Why doesn't Euler's theorem provide an *absolute* energy scale we know that for a one component system
$$U= TS - PV + N \mu$$
Now all terms on the right hand side have well defined zeros, but the internal energy is known to be a relative quantity, with no unique zero. What's going on?
Similarly, we derive the above from the fact that:
$$U(kS, kV, kN)= kU(S,V,N)$$
And it seems to me that taking $k=0$ gives us $U(0,0,0)=0$ - the internal energy of an empty system is the uniquely defined zero of energy.
 A: The statement "the internal energy is known to be a relative quantity, with no unique zero" refers probably to the fact that for common systems of thermodynamics that do not lose or gain matter, the First law of thermodynamics implies the internal energy is a function of temperature $T$ and volume $V$ such that addition of a constant independent of them has no consequence on the measurable quantities; the function
$$
U'(T,V) = U(T,V) + c~~~(1)
$$
is just as good as the function $U(T,V)$ and so for given thermodynamic state of the system, no value of internal energy is preferred. 

Note that neither the function $U$ nor $c$ have $N$ as an argument here because the system does not change its molar number $N$. The relation 
$$
U = TS -PV + \mu N
$$
has no motivation here because $N$ does not appear at all.

However, when considering two or more systems of the same chemical species with the same temperature and pressure but different amount of matter, it is quite natural to assume that for given $T,P$ the internal energy is directly proportional to the molar number $N$ (additivity of energy). One introduces new function $\tilde{U}$ for it that expresses this:
$$
\tilde{U}(T,P,N) = N\tilde{U}(T,P,1)=N\tilde{u}(T,P),~~~(2)
$$
where $\tilde{u}$ is energy of one mole.
This does not follow from the 1st law of thermodynamics, but probably is an independent assumption that is very often true(it may not be true in some cases, for example when capillary forces play role.) This additivity together with other ideas about the entropy then lead to the relation $U=TS-PV+\mu N$.
This relation, however, does not mean that for given thermodynamic state, single value of the internal energy has been found; it only means that there is a relation between energy, entropy and chemical potential. The only quantity that is fixed in this equation is
$$
PV =  TS - U +\mu N
$$
since both $P$ and $V$ are measurable quantities, but entropy, energy or chemical potential are not. One can redefine at least two of them to give the same product $PV$ and other quantities.

In the rest of the post I will show that arbitrariness of $U$ translates into arbitrariness of $\tilde{u}$. For system with constant $N$, since we have two functions $U,U'$, let us assume there are corresponding two functions $\tilde{u},\tilde{u}'$ that give the appropriate values of energy for given state:
$$
U(T,V) = N\tilde{u}(T,P)
$$
$$
U'(T,V) = N\tilde{u}'(T,P)
$$
Combining the two equations (1),(2), we obtain
$$
N\tilde{u}'(T,P) = N\tilde{u}(T,P) + c
$$
The only way this can be true for system of any size is if the quantity $c$ is proportional to $N$. So actually $c$ is not always constant; it is constant only for systems that do not change their $N$. Generally
$$
c=\alpha N
$$
so the energy of one mole
$$
\tilde{u}'(T,P) = \tilde{u}(T,P)+\alpha
$$
is determined only up to an arbitrary constant $\alpha$.

A: I think in your arguments you're neglecting the potential energy from intermolecular forces, or should I say the chemical bonding energy of your system or even mass-energy of the atoms. Meaning that even if you consider the 0 Kelvin case and freeze all degrees of freedom (vibrational, transnational, rotational) you still cannot expect the internal energy $U$ to be zero.  This is taken care of by the internal chemical potential of each molecule, as $\mu$ is composed of two parts: 
External $\mu_{\rm ext}$: electric potential, gravitational, ...
Internal $\mu_{\rm int}$: obtained from molar Gibbs free energy, contains all the dependencies on density, temperature, electrochemical potential, etc.
Finally your argument with coefficient $k$, if fulfilled, just shows that $U$ is extensive with respect to $(S,V,N)$, and does not really contribute to the problem at hand here.
Last but no least, even at $S=0$ i.e. a single microstate available to the system, things do not simplify that much as there is still residual motion in the system due to the uncertainty principle in the ground state.
A: From a thermodynamic perspective, the measurable quantities (specific heat, chemical potential, etc.) are defined as partial derivatives of $U$ with respect to some other variable which is uniquely defined (volume, number of particles, etc.). Thus, changing $U$ by a constant does not change any of the physically measurable quantities.
This is like adding a total time derivative to the Lagrangian, or making a gauge transformation to the electromagnetic potentials. Nothing you can physically measure will change if you make these transformations.
